Solution

$$ \begin{equation*} x^{2}-2 x-8=(x-4)(x+2) \tag{i} \end{equation*} $$

The value of $x^{2}-2 x-8$ is zero when $x-4=0$ or $x+2=0$, i.e., when $x=4$ or $x=-2$

Therefore, the zeroes of $x^{2}-2 x-8$ are 4 and -2 .

Sum of zeroes $=4-2=2=\dfrac{-(-2)}{1}=\dfrac{-(\text{ Coefficient of } x)}{\text{ Coefficient of } x^{2}}$

Product of zeroes $=4 \times(-2)=-8=\dfrac{(-8)}{1}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}}$

(ii) $4 s^{2}-4 s+1=(2 s-1)^{2}$

The value of $4 s^{2}-4 s+1$ is zero when $2 s-1=0$, i.e., $\quad s=\dfrac{1}{2}$

Therefore, the zeroes of $4 s^{2}-4 s+1$ are $\dfrac{1}{2}$ and $\dfrac{1}{2}$.

Sum of zeroes $=\dfrac{1}{2}+\dfrac{1}{2}=1=\dfrac{-(-4)}{4}=\dfrac{-(\text{ Coefficient of } s)}{(\text{ Coefficient of } s^{2})}$

Product of zeroes $=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } s^{2}}$

$$ \begin{equation*} 6 x^{2}-3-7 x=6 x^{2}-7 x-3=(3 x+1)(2 x-3) \tag{iii} \end{equation*} $$

The value of $6 x^{2}-3-7 x$ is zero when $3 x+1=0$ or $2 x-3=0$, i.e., $x=\dfrac{-1}{3}$ or $x=\dfrac{3}{2}$

Therefore, the zeroes of $6 x^{2}-3-7 x$ are $\dfrac{-1}{3}$ and $\dfrac{3}{2}$.

Sum of zeroes $=\dfrac{-1}{3}+\dfrac{3}{2}=\dfrac{7}{6}=\dfrac{-(-7)}{6}=\dfrac{-(\text{ Coefficient of } x)}{\text{ Coefficient of } x^{2}}$

Product of zeroes $=\dfrac{-1}{3} \times \dfrac{3}{2}=\dfrac{-1}{2}=\dfrac{-3}{6}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } x^{2}}$

(iv) $4 u^{2}+8 u=4 u^{2}+8 u+0$

$=4 u(u+2)$

The value of $4 u^{2}+8 u$ is zero when $4 u=0$ or $u+2=0$, i.e., $u=0$ or $u=-2$

Therefore, the zeroes of $4 u^{2}+8 u$ are 0 and -2 .

Sum of zeroes $=0+(-2)=-2=\dfrac{-(8)}{4}=\dfrac{-(\text{ Coefficient of } u)}{\text{ Coefficient of } u^{2}}$

Product of zeroes $=0 \times(-2)=0=\dfrac{0}{4}=\dfrac{\text{ Constant term }}{\text{ Coefficient of } u^{2}}$

(v) $t^{2}-15$

$ \begin{aligned} & =t^{2}-0 t-15 \\ & =(t-\sqrt{15})(t+\sqrt{15}) \end{aligned} $

The value of $t^{2}-15$ is zero when $t-\sqrt{15}=0$ or $t+\sqrt{15}=0$, i.e., when

Solution

(i)

$ \dfrac{1}{4},-1 $

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and $\beta$. $\alpha+\beta=\dfrac{1}{4}=\dfrac{-b}{a}$

$\alpha \beta=-1=\dfrac{-4}{4}=\dfrac{c}{a}$

If $a=4$, then $b=-1, c=-4$

Therefore, the quadratic polynomial is $4 x^{2}-x$ - 4 .

(ii) $\sqrt{2}, \dfrac{1}{3}$

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha+\beta=\sqrt{2}=\dfrac{3 \sqrt{2}}{3}=\dfrac{-b}{a}$

$\alpha \beta=\dfrac{1}{3}=\dfrac{c}{a}$

If $a=3$, then $b=-3 \sqrt{2}, c=1$

Therefore, the quadratic polynomial is $3 x^{2}-3 \sqrt{2} x+1$.

(iii) $0, \sqrt{5}$

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha+\beta=0=\dfrac{0}{1}=\dfrac{-b}{a}$

$\alpha \times \beta=\sqrt{5}=\dfrac{\sqrt{5}}{1}=\dfrac{c}{a}$

If $a=1$, then $b=0, c=\sqrt{5}$

Therefore, the quadratic polynomial is $x^{2}+\sqrt{5}$.

(iv) 1,1

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha+\beta=1=\dfrac{1}{1}=\dfrac{-b}{a}$

$\alpha \times \beta=1=\dfrac{1}{1}=\dfrac{c}{a}$

If $a=1$, then $b=-1, c=1$

Therefore, the quadratic polynomial is $x^{2}-x+1$.

(v) $-\dfrac{1}{4}, \dfrac{1}{4}$

Let the polynomial be $a x^{2}+b x+c$, and its zeroes be $\alpha$ and